A pendulum is swinging back and forth. After $t$ seconds, the horizontal distance from the bob to the place where it was released is given by $ H(t) = 7 - 7 \cos \left(\dfrac{2\pi (t-2)}{20}\right)$. How often does the bob cross its midline? Give an exact answer Every
The bob crosses its midline whenever $\cos \left(\dfrac{2\pi (t-2)}{20}\right) = 0$. Since $\cos \theta = 0$ when $\theta = \pm \dfrac{\pi}{2} + 2\pi n$, we can find then the bob crosses its midline by solving $\begin{aligned} \dfrac{2\pi (t-2)}{20}&= \pm \frac{\pi}{2} + 2\pi n\\ \\ t - 2 &= \pm 5 + 20 n.\\ \end{aligned}$ The solutions are when $t = -3, 7, 17, 27, 37,...$ The bob passes its midline every $10$ seconds.